Here's an outline of the standard proof of the Spectral Theorem for self-adjoint oprators T on a finite-dimensional real inner product space V. This is clearer than Axler's approach (pp.134 ff.), and does not depend on the Fundamental Theorem of Algebra. Since I don't know how to get a ``lambda'' or ``epsilon'' in HTML (are there commands for these?), I'll use c and e instead.

Our aim is to show that there is an orthonormal basis v1, v2, ..., vn for V such that Tvj = cjvj for each j. It is enough to prove the existence of a single eigenvector, i.e. of a nonzero vector v1 such that Tv1 = c1v1 for some real number c1. This is because (using the fact that T=T*) if v is an eigenvector and <v,v'>=0 then <Tv,v'>=0. Thus T may be regarded as a self-adjoint operator on the orthogonal complement of v1; if that complement is not the zero space, it will then contain an eigenvector v2 by the same argument, which we then repeat for the orthogonal complement of the span of v1 and v2, and so on until we obtain our basis. (This concluding part of the argument is done by Axler (see 7.13); the difference is how we get v1 in the first place.)

Assume that we have succeeded. Without loss of generality we may list the eigenvalues cj in decreasing order. Then for all v in V we find that <Tv,v> is at most c1 <v,v>, with equality if and only if Tv = c1v. The ratio between <Tv,v> and <v,v> is called the Rayleigh quotient RT(v). [Of course we must have v nonzero.] Note that RT(v) = RT(av) for any nonzero scalar a. This suggests the following strategy: Find v1 which maximizes RT(v) subject to <v,v>=1, and show that this v1 is an eigenvector.

The point is that since V is finite-dimensional, the function from V to R taking any v to <Tv,v> is continuous (it's just a quadratic polynomial in the coordinates), and the unit sphere {v : <v,v>=1} is compact (using Heine-Borel and the continuity of the function taking v to <v,v>). Now, on <v,v>=1, the Rayleigh quotient simplifies to <Tv,v>. So this function is bounded and attains its maximum. Call that maximum c1, and let v1 be a vector that attains it, i.e. a vector such that <v1,v1>=1 <Tv1,v1>=c1.

It remains to show that Tv1=c1v1. Let u be any vector in V. Then <T(v1+eu),v1+eu> is at most c1<v1+eu,v1+eu> for any scalar e. The difference is a linear combination of e and e2; for it to be nonnegative for all e, the e coefficient must vanish. This coefficient is 2<Tv1-c1v1,u>. Taking u=Tv1-c1v1 we find that Tv1-c1v1=0. Therefore Tv1=c1v1 as desired.